I had written down so many times, and kept trying to wrong things with it. Is it as simple as this?:

is a sum of two squares which is prime. And because a composite is the product of two primes, is composite?

it is, as topsquark says, even easier than that. you have an integer that is three times another integer. it means that your integer is divisible by 3 (an integer n is divisible by 3 if it can be written in the form for some integer ), and so is not prime.

This means that, for any n, is divisible by 3 and is thus a composite number.

Originally Posted by Jhevon

it is, as topsquark says, even easier than that. you have an integer that is three times another integer. it means that your integer is divisible by 3 (an integer n is divisible by 3 if it can be written in the form for some integer ), and so is not prime.

Not true! 3 is divisible by 3 but 3 is not a composite number!

A positive integer divisible by 3 is not prime only if it’s not equal to 3 itself. It this particular problem, is not prime because 3 divides it and is not equal to 1 for any integer n.

I had to learn the hard way too. Once I wanted to prove that, while twin primes exist, you could never have “triple primes” (three consecutive odd primes). I thought I had cool proof in just one line: “Any sequence of three consecutive odd integers must contain a multiple of 3 and so that number in the sequence cannot be prime.” Then someone knocked me off my pedestal by showing me that one set of triple primes does exist: 3, 5, 7. I have learnt my lesson since then!